# Pendulum and constraining forces (Lagrangian mechanics)

1. May 12, 2016

### JulienB

1. The problem statement, all variables and given/known data

Hi everybody! As always, I struggle with my special relativity class and here is a new problem I'd like to have some indications about:

A masspoint m moves in the x-y-plane under the influence of gravity on a circular path of radius r (see attached pic). Which constraining force $\vec{z}$ must be additionally be acting in the radial direction, so that the masspoint remains on the circular path? You must first establish the equation of motion for the masspoint m and integrate it for the case small angle $\varphi$.

2. Relevant equations

Lagrangian mechanics, holonomic constraints

3. The attempt at a solution

So first I wrote the coordinates dependently of each other and in function of $l$:

$x^2 + y^2 = l^2 \iff x^2 + y^2 - l^2 = 0$

If I understand my script correctly, such an equation $f(x,y,l,t) = 0$ means this is a case of holonomic constraint. Right? I am not sure if $l$ should be included in my function. Anyway I can rewrite $x$ and $y$ as

$x = l \cdot \cos \varphi$ and $y = l \ cdot \sin \varphi$
$\implies \dot{x} = - \varphi l \sin \varphi$ and $\dot{y} = \varphi l \cos \varphi$

I can now substitute those values in the Lagrange function to have an equation of motion only depending on $l$ and $\varphi$:

$L = T - V = \frac{1}{2} m (\dot{x}^2 + \dot{y}^2) - mgh$
$= \frac{1}{2} m\varphi^2 l^2 (\cos^2\varphi + \sin^2\varphi) - mgh$
$= \frac{1}{2} m \varphi^2 l^2 - mgh$

That went pretty well until now, but now I'm blocked because of the $h$. How can I express it in terms of $l$ and $\varphi$? Geometrically I see the relation but I don't know what $h$ is. Anything I am missing?

Julien.

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2. May 12, 2016

### BvU

Wouldn't $\left (1 - \cos\phi\right)$ qualify for the job ?

3. May 12, 2016

### JulienB

@BvU You mean for $h$? Did you see the picture? I searched a bit on the internet, and everyone seems to use a different coordinate system. For example in page 5 of this link (http://www.math.pitt.edu/~bard/classes/1270/mechanics.pdf) is $h=-l\cos \varphi$ apparently.

4. May 12, 2016

### JulienB

(Note that my teacher has inversed the $x$ and $y$ axes, probably to make it just confusing enough so that we don't get it.. I'm unsure why is $y = -l \cos \varphi$ in the link I put in post #3 and not $l \cos \varphi$ as for my $x$. Is that just because my $x$ axis points downwards?)

5. May 12, 2016

### JulienB

I think I got it, I must have misunderstood something about the potential. Apparently I can just define a zero potential energy point which conveniently would be $\varphi = 0$, then my potential would be as said @BvU $V = -mg\Delta h = -mg l (1 - cos \varphi)$. Is that correct?

6. May 12, 2016

### BvU

Yes, potential can be defined wrt an arbitrary point.
And x/y are now out of the picture with this generalized coordinate, so which way they point isn't all that relevant anymore. But I agree the choice is somewhat unusual.

7. May 13, 2016

### JulienB

@BvU Okay thanks a lot :) Then I get an equation of motion $m \ddot{\varphi} = \frac{-m g \cdot \varphi}{l}$. If I am not mistaken, one can find the constraint force from the equation of motion with $\sum \vec{F} = \vec{F_G} + \vec{Z}$, right?

Does that mean $\vec{Z} = \frac{-m g \cdot \varphi}{l} - mg$?

Thanks a lot.

Julien.

Last edited: May 13, 2016